I'm going to sub in z(x)=x-y. Then dx=dz and f(x-y)=f(z) etc. Thanks for the help guys :) This looks on the right track according to some of the proofs I've seen.
I don't think that matters. It's a function of the variable (x-y) not y. Can you not sub in z=x-y? Then you no longer have that problem. I don't think it'll fall under the dy integral for that reason since it's not a function of the y variable.
I can't actually solve the integral as I'm supposed to be subbing in the symbol for the Fourier transform. I have no choice but to separate them somehow.
Here's the prof's solution. I'm not sure how he got to the 3rd step though. How did he get d(x-y) out of dx...
Show G(k)=\sqrt{2π}g1(k)g2(k)
Given that G(k) is the Fourier transform of F(x), g1(k) is Fourier trans of f1(x), g2(k) is Fourier trans of f2(X) and
F(x)=^{+∞}_{-∞}∫dyf1(y)f2(x-y)
SO FAR
G(k)=1/\sqrt{2π}^{+∞}_{-∞}∫F(x)e-ikxdx <-def'n of Fourier transform...